A Quantum Quench of the Sachdev-Ye-Kitaev Model Julia Steinberg - - PowerPoint PPT Presentation

A Quantum Quench of the Sachdev-Ye-Kitaev Model

Julia Steinberg Harvard University arXiv:1703.07793 [cond-mat.str-el] Chaos, Topology, and Dualities in Condensed Matter Theory UIUC November 4, 2017

Andreas Eberlein Harvard University Valentin Kasper Harvard University Subir Sachdev Harvard University Perimeter Institute of Theoretical Physics

Collaborators

Quantum matter without quasiparticles

• Want to study properties of systems without quasiparticles
• First: what is a quasiparticle?
• Long lived additive excitation with same quantum numbers as free particle
• How do we identify systems without quasiparticles?
• Fastest relaxation
• No long lived excitations in any basis
• “Too fast”: cannot study long time behavior with conventional techniques

τeq ≥ C ~ kBT , T → 0

The SYK model: a solvable system without quasiparticles

• Model of N flavors of Majorana fermions with infinite range q-body interactions
• Solvable in large N limit
• Maximally chaotic
• Disorder average→melon diagrams, only keep one
• Full propagator

H = (i)

q 2

X

1≤i1<i2<...<iq≤N

ji1i2...iqψi1ψi2...ψiq hj2

i1...iqi = J2(q 1)!

N q−1 . = +

Strong coupling behavior

• For →“Emergent reparameterization symmetry
• Spontaneously and explicitly broken→ Schwarzian action
• SYK is “dual” to nearly

βJ 1

↔

iGR(t) = C(J, ∆) 1 β sinh πt

β

!2∆ θ(t) ∆ = 1 q AdS2

Escher “Heaven and hell”

Quantum quenches and thermalization

• Quench of SYK→ probe nonequilibrium dynamics without quasiparticles
• Possibly reach thermal state
• What is a thermal state?
• System is its own heat bath for subsystems
• Steady state, observables reach thermal values
• For SYK, our working definition of thermalization:

2-point function obeys KMS, T from energy conservation

B A

Quench procedure

• Start with SYK model with q and pq interactions
• Turn off pq term instantaneously
• Track evolution of Green’s function
• Does SYK Greens function thermalize?
• How long does it take (scaling dependence on T)?
• What is the best way to do this?

Green’s Functions on the Closed-Time- Contour

• Out of equilibrium, must study full evolution along contour
• Two Greens functions: and
• Use to form 2 by 2 matrix
• Dyson (matrix) equation from disorder average:

Σ(t1, t2) = X

i

iqiJ2

qiG(t1, t2)qi−1

G−1

0 (t1, t2) − G−1(t1, t2) = Σ(t1, t2)

G>(t1, t2) ≡ G(t−

1 , t+ 2 )

G<(t1, t2) ≡ G(t+

1 , t− 2 )

The Kadanoff-Baym equations

• How do we study 2-pt function with no time translation?
• Kadanoff-Baym equation directly from Dyson equation
• Go to real time plane get two integro-differential equations:
• For Majorana fermions have condition:
• Always true→everything from !

G−1 ⊗ G> =

• ΣR ⊗ G> + Σ> ⊗ GA

G> ⊗ G−1 =

• GR ⊗ Σ> + G> ⊗ ΣA

G>(t1, t2) = −G<(t2, t1) G>(t1, t2) =

Causal Structure

• Evolution from integral structure of Kadanoff Baym equations
• Rewrite everything in terms of
• Step functions→limits of integration
• For point → integrate “rectangle region”
• Pre-quench
• Post quench
• Pass through other quadrants→causal effect

G>(t1, t2) =

>(t1, t2) =

t2 t1

>(t1, t2) =

t1 ≤ 0, t2 ≤ 0 t1 ≥ 0, t2 ≥ 0

Numerics model

• Consider the SYK+random matrix model (p=1/2 q=4)
• Only pq: integrable
• Both terms: “Fermi liquid”
• Only q: “strange metal”
• Use and to specify quench
• Solve full Kadanoff-Baym equations numerically
• Use Majorana condition → solve for

H(t) = i X

i<j

j2,ijf(t)ψiψj − X

i<j<k<l

j4,ijklg(t)ψiψjψkψl τ −1

eq ∼ T

τ −1

eq ∼ T 2 ,ijf(t)ψ klg(t)ψ

G>(t1, t2) =

Procedure

• Solve Dyson equation self-consistently for thermal initial state
• Use as BC for quench
• Immediately post quench, no time translation invariance, define
• Absolute time:
• Relative time:
• Near equilibrium, varies slowly with look at low frequency behavior
• Wigner transform
• Also define

T = t1 + t2 2 t = t1 − t2 f(t1, t2) → f(T , ω) T GK(t1, t2) = G>(t1, t2) + G<(t1, t2)

Thermal state from the KMS condition

• “Thermal” 2-pt function obeys KMS
• KMS→FDT
• “Effective inverse T”:
• Start with thermal state
• Right after quench out of equilibrium
• , varies slowly→“thermal”

−2.5 2.5 ω/J4,f −1 1 iGK(T , ω)/A(T , ω) T J4 = −50 T J4 = 0 T J4 = 50 J2,i = 0.5, J2,f = 0, J4,i = J4,f = 1, Ti = 0.04J4

iGK(T , ω) A(T , ω) = tanh ✓β(T )ω 2 ◆ β(T ) T → ∞ β(T )

Effective temperature

• from fit of FDT relation
• Relaxes exponentially
• Check throughout quench
• Determines
• Depends on only through

−50 50 T 0.05 0.075 0.1 Teff Ti = 0.04J4 Ti = 0.08J4 J2,i = 0.0625, J2,f = 0, J4,i = J4,f = 1

Teff hHi = Ef Teff J2 Ef

Relaxation rate

• Know final temperature from energy

conservation

• How long does it take to reach final

temperature for ?

• Exponential rate
• Higher temperature, controlled by

βJ 1 Γ ∝ T Γ ∝

The large q limit

• Consider large q interaction (after Large N)
• Expand
• exponential form of self energy
• Derivatives of KB eqns → Lorentzian-Liouville eqn
• Exact solution for p=1/2, or 2

J 2(t) = qJ2(t)21−q , J 2

p (t) = qJ2 p(t)21−pq

q → ∞

G>(t1, t2) = i hψ(t1)ψ(t2)i = i 2  1 + 1 q g(t1, t2) + . . .

• ∂2

∂t1∂t2 g(t1, t2) = 2J (t1)J (t2)eg(t1,t2) + 2Jp(t1)Jp(t2)epg(t1,t2) G>(t1, t2) =

Quench Regions

• General solution in all regions
• Majorana condition →
• For equilibrium solution
• Structure of integrals in KB equations show this is always true
• Need to solve in 5 regions

g(t, t) = 0 g(t2, t1) = [g(t1, t2)]∗ g(t1, t2) = ln " −h

1(t1)h 2(t2)

J 2(h1(t1) − h2(t2))2 # t1 ≤ 0, t2 ≤ 0

Quench time plane

∂ ∂t1 g(t1, t2) = 2 Z t2

−∞

dt3 J (t1)J (t3)eg(t1,t3)− Z t1

−∞

dt3 J (t1)J (t3) h eg(t1,t3) + eg(t3,t1)i +2 Z t2

−∞

dt3Jp(t1)Jp(t3)epg(t1,t3) − Z t1

−∞

dt3Jp(t1)Jp(t3) h epg(t1,t3) + epg(t3,t1)i ∂ ∂t2 g(t1, t2) = 2 Z t1

−∞

dt3 J (t3)J (t2)eg(t3,t2)− Z t2

−∞

dt3 J (t3)J (t2) h eg(t3,t2) + eg(t2,t3)i +2 Z t1

−∞

dt3Jp(t3)Jp(t2)epg(t3,t2) − Z t2

−∞

dt3Jp(t3)Jp(t2) h epg(t3,t2) + epg(t2,t3)i

Boundary Conditions

• Need 3 BCs
• Get , and from , , ,
• Too many BCs
• SL(2,C) invariance:
• Constraint:
• Result independent of choices for , ,
• SL(2,R) invariance→ “gauge” choice!

h(t) → a h(t)+b

c h(t)+d

ad − bc = 1 hA1(0) hA2(0) h0

A1(0)

h0

B2(0)

hB2(0) hB1(−∞) gC(t) hB2(0) hB1(−∞) h0

B2(0)

Post-Quench Solution

• Choose ansatz
• Find solution for p=1/2
• From KMS:
• Only depends on relative time: instant thermalization!

gA(t1, t2) = ln  −σ2 4J 2 sinh2(σ(t1 − t2)/2 + iθ)

• hA1(t) = aeσt + c

ceσt + d , hA2(t) = ae−2iθeσt + b ce−2iθeσt + d βf = 2(π − 2θ) σ σ = 2J sin θ e−4iθ = (b − dhB1(−∞))(a∗ − c∗h∗

B1(−∞))

(b∗ − d∗h∗

B1(−∞))(a − chB1(−∞))

Relation to the Schwarzian Action

• SYK described by Schwarzian for
• Take from KB equations
• is solution to Schwarzian EOM
• Thermalization connected to reparameterization modes
• Schwarzian should also exhibit instantaneous thermalization

[h0(t)]2 h0000(t) + 3 [h00(t)]3 − 4h0(t)h00(t)h000(t) = 0 βJ 1 L[h(t)] = h000(t) h0(t) − 3 2 ✓h00(t) h0(t) ◆2

h(t) h(t)

Final Remarks

• Low energy limit, rate linear in T as expected
• Also depends on q, what do 1/q2 corrections look like?
• 2-point function instantly thermalizes, but other quantities do not
• Which quantities thermalize and on what timescale (some do not)?
• When does the large N limit break down?
• What other consequences does this have in gravity?